SOLUTION: Combustion analysis pogil ap chem - Studypool - Free Printable

It looks like you've uploaded a page from a chemistry worksheet or textbook focused on combustion analysis of hydrocarbons (CₓHᵧ). The goal is to determine the empirical formula of an unknown hydrocarbon based on the masses of CO₂ and H₂O produced during combustion.

Let’s go through Model 2 – Combustion Analysis of CₓHᵧ Unknowns, fill in any missing values, verify the calculations, and explain how to solve such problems step by step.

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🔍 Problem Overview:

In combustion analysis:
- A hydrocarbon burns in excess O₂.
- All carbon → CO₂
- All hydrogen → H₂O
- We measure mass of CO₂ and H₂O produced.
- From that, we find moles of C and H in the original sample.
- Then determine empirical formula.

We are given:

| Sample | Mass of CO₂ | Moles of CO₂ | Moles of C atoms | Mass of H₂O | Moles of H₂O | Moles of H atoms | Empirical Formula | Total Mass of C & H |
|--------|-------------|--------------|------------------|-------------|--------------|------------------|--------------------|---------------------|
| 1 | 27.42 g | ? | ? | 22.46 g | ? | ? | CH₄ | ? |
| 2 | 29.26 g | ? | ? | 17.97 g | ? | ? | CH₃ | ? |

We’ll complete this table and verify the work.

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## ✔ Step-by-Step Solution for Each Sample

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🔹 Sample 1: 10.00-g sample

#### Given:
- Mass of CO₂ = 27.42 g
- Mass of H₂O = 22.46 g

#### Step 1: Moles of CO₂
Molar mass of CO₂ = 44.01 g/mol
$$
\text{Moles of CO₂} = \frac{27.42}{44.01} = 0.6230 \text{ mol}
$$

✔ Matches your handwritten value: 0.6230 mol

#### Step 2: Moles of Carbon Atoms
Each mole of CO₂ contains 1 mole of C
So,
$$
\text{Moles of C} = 0.6230 \text{ mol}
$$

✔ Matches: 0.6230 mol

#### Step 3: Mass of Carbon
$$
\text{Mass of C} = 0.6230 \times 12.01 = 7.482 \text{ g}
$$

(We'll use this later)

#### Step 4: Moles of H₂O
Molar mass of H₂O = 18.02 g/mol
$$
\text{Moles of H₂O} = \frac{22.46}{18.02} = 1.2463 \text{ mol}
$$

✔ Matches: 1.2463 mol

#### Step 5: Moles of Hydrogen Atoms
Each mole of H₂O has 2 moles of H
$$
\text{Moles of H} = 1.2463 \times 2 = 2.4926 \text{ mol}
$$

✔ Matches: 2.4926 mol

#### Step 6: Mass of Hydrogen
$$
\text{Mass of H} = 2.4926 \times 1.008 = 2.512 \text{ g}
$$

#### Step 7: Total Mass of C and H
$$
7.482 + 2.512 = 9.994 \approx 10.00 \text{ g}
$$

✔ So, all carbon and hydrogen accounted for — no oxygen present (pure hydrocarbon).

#### Step 8: Find Empirical Formula

We have:
- C: 0.6230 mol
- H: 2.4926 mol

Divide both by smallest:
$$
\text{C: } \frac{0.6230}{0.6230} = 1 \\
\text{H: } \frac{2.4926}{0.6230} \approx 4.00
$$

So, ratio is C₁H₄ → CH₄

✔ Matches: CH₄

#### Step 9: Total Mass of C and H Atoms
As calculated: ~10.00 g

✔ You wrote 16.05 g — that’s wrong.

Wait! That must be a mistake.

Wait — 16.05 g? But total sample is only 10.00 g.

So either:
- You meant mass of C and H atoms in the sample, which should be ~10.00 g
- Or there's confusion with molar mass.

But 16.05 g is close to molar mass of CH₄ (12.01 + 4×1.008 = 16.04), but that’s not what’s asked.

> ✘ "Total Mass of C and H Atoms" means mass of C and H in the sample, not molar mass.

So correct answer: ≈10.00 g

✔ So your 16.05 g is incorrect. It should be ~10.00 g.

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🔹 Sample 2: 10.00-g sample

#### Given:
- Mass of CO₂ = 29.26 g
- Mass of H₂O = 17.97 g

#### Step 1: Moles of CO₂
$$
\frac{29.26}{44.01} = 0.6647 \text{ mol}
$$

✔ Matches: 0.6647 mol

#### Step 2: Moles of C atoms = 0.6647 mol

#### Step 3: Mass of C
$$
0.6647 \times 12.01 = 7.982 \text{ g}
$$

#### Step 4: Moles of H₂O
$$
\frac{17.97}{18.02} = 0.9972 \text{ mol}
$$

✔ Matches: 0.9972 mol

#### Step 5: Moles of H atoms
$$
0.9972 \times 2 = 1.9944 \text{ mol}
$$

✔ Matches: 1.994 ≈ 1.994 mol

#### Step 6: Mass of H
$$
1.9944 \times 1.008 = 2.008 \text{ g}
$$

#### Step 7: Total Mass of C and H
$$
7.982 + 2.008 = 9.990 \approx 10.00 \text{ g}
$$

✔ Again, accounts for entire sample.

#### Step 8: Empirical Formula

C: 0.6647 mol
H: 1.9944 mol

Divide by smallest:
$$
\text{C: } \frac{0.6647}{0.6647} = 1 \\
\text{H: } \frac{1.9944}{0.6647} \approx 3.00
$$

→ CH₃

✔ Matches: CH₃

But wait — CH₃ is not a stable molecule. It’s a radical.

That means the empirical formula is CH₃, but the actual molecular formula must be a multiple (like C₂H₆, C₃H₉, etc.).

So CH₃ is acceptable as empirical formula.

#### Step 9: Total Mass of C and H Atoms
Again, ~10.00 g

You wrote 15.04 g — again, likely a mistake.

Why 15.04?

If you multiply molar mass of CH₃ (12.01 + 3×1.008 = 15.03) — you’re confusing molar mass with mass in sample.

✘ So 15.04 g is incorrect.

Correct: ~10.00 g

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✔ Corrected Table

| 10.00-g Sample | Mass of CO₂ | Moles of CO₂ | Moles of C atoms | Mass of H₂O | Moles of H₂O | Moles of H atoms | Empirical Formula | Total Mass of C and H Atoms |
|----------------|-------------|--------------|------------------|-------------|--------------|------------------|--------------------|------------------------------|
| 1 | 27.42 g | 0.6230 mol | 0.6230 mol | 22.46 g | 1.2463 mol | 2.4926 mol | CH₄ | 10.00 g |
| 2 | 29.26 g | 0.6647 mol | 0.6647 mol | 17.97 g | 0.9972 mol | 1.9944 mol | CH₃ | 10.00 g |

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🚨 Key Takeaway:

- The total mass of C and H atoms in the sample equals the mass of the original sample if it’s a pure hydrocarbon (no O).
- Your written values (16.05 g, 15.04 g) are molar masses of CH₄ and CH₃, not the mass of elements in the sample.
- So those entries are incorrect.

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💡 How to Solve Any Combustion Problem:

1. Find moles of CO₂ → moles of C
$$
\text{mol C} = \frac{\text{mass CO}_2}{44.01}
$$

2. Find moles of H₂O → moles of H
$$
\text{mol H} = 2 \times \frac{\text{mass H}_2\text{O}}{18.02}
$$

3. Convert moles to mass (optional):
- Mass C = mol C × 12.01
- Mass H = mol H × 1.008

4. Add mass C + mass H → should equal sample mass (if no O)

5. Find simplest whole number ratio:
- Divide moles of C and H by smallest value
- Round to nearest whole number

6. Write empirical formula

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✔ Final Answer Summary:

- Sample 1: Empirical formula = CH₄, total mass of C and H = 10.00 g
- Sample 2: Empirical formula = CH₃, total mass of C and H = 10.00 g

⚠️ Do not confuse molar mass of empirical formula with actual mass of elements in sample.

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Let me know if you want help with other samples or questions below the table (they’re cut off).